Mid-Semester Quiz Second Semester, 2015

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1 THE AUSTRALIAN NATIONAL UNIVERSITY Mid-Semester Quiz Second Semester, 2015 COMP2600/6260 (Formal Methods for Software Engineering) Writing Period: 1 hour duration Study Period: 0 minutes duration Permitted Materials: One A4 page with hand-written notes on both sides The questions are followed by labelled blank spaces into which your answers are to be written. Additional answer panels are provided at the end of the paper should you wish to use more space for an answer than is provided in the associated labelled panels. Student Number: Q1 Mark Q2 Mark Q3 Mark Total Mark COMP2600/6260 (Formal Methods for Software Engineering) Page 1 of 14

2 QUESTION 1 [10 marks] Structural Induction Here is the usual Haskell definition of a binary tree: data Tree a = Nul Node a ( Tree a) ( Tree a) Given these function definitions: vertices :: Tree a - > Int vertices Nul = 1 -- V1 vertices ( Node x l r) = 1 + ( vertices l) + ( vertices r) -- V2 edges :: Tree a - > Int edges Nul = 0 -- E1 edges ( Node x l r) = 2 + ( edges l) + ( edges r) -- E2 Prove by structural induction that the following property holds for all trees t of type Tree a: vertices t = edges t + 1 (a) State and prove the base case goal. QUESTION 1(a) [2 marks] Base case: t = Nil Show that vertices t = edges t + 1 Proof: vertices Nul 1 -- V arithmetic = edges Nul E1 COMP2600/6260 (Formal Methods for Software Engineering) Page 2 of 14

3 (b) State the induction hypotheses. QUESTION 1(b) [2 marks] For proving the step case for t = (Node x l r), the induction hypotheses are vertices l = edges l IH1 vertices r = edges r IH2 (c) State and prove the step case goal. QUESTION 1(c) [6 marks] Step case: t = (Node x l r) Show that if the inductive hypotheses (as above) holds, then vertices (Node x l r) = edges (Node x l r) + 1 Proof: vertices (Node x l r) = 1 + (vertices l) + (vertices r) -- V2 = 1 + (edges l + 1) + (edges r + 1) -- IH1 and IH2 = 2 + (edges l) + (edges r) arithmetic = edges (Node x l r) E2 COMP2600/6260 (Formal Methods for Software Engineering) Page 3 of 14

4 QUESTION 2 [14 marks] Logic (a) Use truth tables to determine whether the following proposition is valid, or is a contradiction, or is a contingency: (p (q r)) (p q) (p r) In addition to giving the truth table, your answer needs to explicitly state whether the proposition is valid, or is a contradiction, or is a contingency. For brevity, the truth table template below uses the abbreviations: R 1 (p q) (p r) R 2 (p (q r)) (p q) (p r) QUESTION 2(a) [2 marks] p q r q r p (q r) p q p r R 1 R 2 T T T T T T T T T T T F T T T F T T T F T T T F T T T T F F F F F F F T F T T T T T T T T F T F T T T T T T F F T T T T T T T F F F F T T T T T The formula R 1 is true for every assignment of truth values to its variables, therefore R 1 is valid. COMP2600/6260 (Formal Methods for Software Engineering) Page 4 of 14

5 (b) Consider the following truth table: p q r A T T T T T T F F T F T T T F F T F T T T F T F T F F T T F F F T Write down a propositional formula A (in variables p, q, r) that behaves as prescribed by the truth table above. (That is, the formula A should evaluate to the truth value given in the right hand column given the truth values of p, q, and r in the three left hand columns). Briefly justify your answer. QUESTION 2(b) [2 marks] One formula is p q r. The only way to make an implication evaluate to false is when the premise p q evaluates to true, and the conclusion r to false. COMP2600/6260 (Formal Methods for Software Engineering) Page 5 of 14

6 (c) Give a natural deduction proof of the following rule: (p q) (q r) p q r You may only use the introduction and elimination rules given in Appendix 1. Number each line and include justifications for each step. QUESTION 2(c) [5 marks] 1 (p q) (q r) 2 p q -E, 1 3 q r -E, 1 4 p q 5 p 6 q -E, 2, 5 7 r -E, 3, 6 8 q 9 r -E, 3, 8 10 r -E, 4, 5 7, (p q) r -I, 4, 10 COMP2600/6260 (Formal Methods for Software Engineering) Page 6 of 14

7 (d) Give a natural deduction proof of the following formula: x.(p(x) ( y.q(y) P(x)) You may only use the introduction and elimination rules given in Appendix 1. Number each line and include justifications for each step. QUESTION 2(d) [5 marks] 1 a P(a) 2 b Q(b) 3 P(a) R, 1 4 Q(b) P(a) -I, 2, 3 5 y.q(y) P(a) -I, 2, 4 6 P(a) ( y.q(y) P(a)) -I, 1, 5 7 x.(p(x) ( y.q(y) P(x)) -I, 1, 6 COMP2600/6260 (Formal Methods for Software Engineering) Page 7 of 14

QUESTION 3 [16 marks] (a) Design a Finite State Automaton that recognises the language of all strings over the alphabet Σ = {a, b, c} where no a is followed by b, and no b is followed by c, and no c

8 QUESTION 3 [16 marks] (a) Design a Finite State Automaton that recognises the language of all strings over the alphabet Σ = {a, b, c} where no a is followed by b, and no b is followed by c, and no c is followed by a. QUESTION 3(a) [5 marks] (b) Is your Finite State Automaton (above) deterministic or non-deterministic? Explain. QUESTION 3(b) [1 mark] It is an NFA. There are several transitions missing. A sink state would be needed for the missing N(S a, b), N(S b, c) and N(S c, a) transitions. COMP2600/6260 (Formal Methods for Software Engineering) Page 8 of 14

9 (c) What language is recognised by the following Finite State Automaton? 0 S S 1 0 S 2 0, 1 Describe the recognised language in English, and give a regular expression defining the language. QUESTION 3(c) [3 marks] It accepts a 0 followed by any number of instances of the string 10. Regular Expression: 0(10) (d) Consider following statement about the automaton given in part (c): w Σ. N (S 2, w) = S 2 Express this statement in English. Why might it be relevant? QUESTION 3(d) [2 marks] From state S 2 any character or string of characters will leave you in state S 2. Once you are in S 2 no transitions allows exit. S 2 is a sink state. It is designed to take all transitions following a rejected string. COMP2600/6260 (Formal Methods for Software Engineering) Page 9 of 14

10 (e) For the Finite State Automaton given in part (c), prove that n N. N (S 1, (10) n ) = S 1. QUESTION 3(e) [5 marks] N (S 1, (10) n ) = S 1 Base Case: n = 0 N (S 1, ɛ) = S 1 (by N1) Step Case: Assume N (S 1, (10) i ) = S 1 - IH Prove N (S 1, (10) (i+1) ) = S 1 RHS = N (S 1, (10) (i+1) ) (1) = N (S 1, 10(10) i ) (2) = N (N(S 1, 1), 0(10) i )(by N2) (3) = N (N(N(S 1, 1), 0), (10) i )(by N2) (4) = N (N(S 2, 0), (10) i )(by N) (5) = N (S 1, (10) i )(by N) (6) = S 1 (by IH) (7) COMP2600/6260 (Formal Methods for Software Engineering) Page 10 of 14

11 Additional answers. Clearly indicate the corresponding question and part. Additional answers. Clearly indicate the corresponding question and part. COMP2600/6260 (Formal Methods for Software Engineering) Page 11 of 14

12 Additional answers. Clearly indicate the corresponding question and part. Additional answers. Clearly indicate the corresponding question and part. COMP2600/6260 (Formal Methods for Software Engineering) Page 12 of 14

13 Appendix 1 Natural Deduction Rules Propositional Calculus ( I) p q p q ( E) p q p p q q [p] [q] ( I) p p q p q p ( E).. p q r r r [p]. ( I) q p q ( E) p q p q [p] [ p].. ( I) q q p ( E) q q p Predicate Calculus ( I) P(a) (a arbitrary) x. P(x) ( E) x. P(x) P(a) [P(a)] ( I) P(a) x. P(x) ( E) x. P(x). q (a arbitrary) q (a is not free in q) COMP2600/6260 (Formal Methods for Software Engineering) Page 13 of 14

14 Appendix 2 Truth Table Values p q p q p q p q p p q T T T T T F T T F T F F F F F T T F T T F F F F F T T T COMP2600/6260 (Formal Methods for Software Engineering) Page 14 of 14

Handout #1: Mathematical Reasoning

Math 101 Rumbos Spring 2010 1 Handout #1: Mathematical Reasoning 1 Propositional Logic A proposition is a mathematical statement that it is either true or false; that is, a statement whose certainty or